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Karolina [17]
2 years ago
13

Which option is an element

Chemistry
1 answer:
Anna35 [415]2 years ago
4 0
Please show the full question!
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What are the missing coefficients for the chemical equation <br> [ ] Ca + [ ] O2 → [ ] CaO
katen-ka-za [31]

Answer:

{ \sf{[ 2] Ca_{(s)} + [1 ] O_{2(g)} → [ 2] CaO _{(s)}}}

4 0
2 years ago
(WILL GIVE BRAINLIEST IF YOU ANSWER ALL 4) Chemistry
Kitty [74]

Answer:

See explanations

Explanation:

a. Molarity = moles/Volume in Liters = 5moles/2Liters = 2.5M in NaCl

b. Freezing Pt Depression

     1. Sprinkling salt on icy surfaces

    2. Using antifreeze in automobile cooling systems

    3. <em>Not an application </em>

    4. Using salt to make ice cream

c. pOH = -log[OHˉ] = -log(1x10ˉ¹⁰) = -(-10) = 10 => pH = 14 – pOH = 14 – 10 = 4

d. H₂O + NH₃ => NH₄⁺ + OHˉ => Bronsted Acid is H₂O  (proton donor)

5 0
3 years ago
8. Calculate the number of moles of eachsubstance.a. 5.45 x 1026 particles of methane, CH4
Klio2033 [76]

<em>ANSWER</em>

The number of moles of methane is 905.32 moles

STEP-BY-STEP EXPLANATION:

Given information

The number of particles of methane = 5.45 x 10^26 particles

Let x represents the number of moles of methane

To calculate the number of moles, we will be using the below formula

\text{Number of particles = number of moles x Avogadro's constant}

Recall that, the Avogadro's constant is given as

6.02\cdot10^{23}\begin{gathered} 5.45\cdot10^{26}\text{ = x }\cdot\text{ 6.02 }\cdot10^{23} \\ \text{Divide both sides by 6.02 }\cdot10^{23} \\ x\text{ = }\frac{5.45\cdot10^{26}}{6.02\cdot10^{23}} \\ x\text{ = }\frac{5.45}{6.02}\cdot10^{26\text{ - 23}} \\ x\text{ = 0.9053 }\cdot10^3 \\ x\text{ = 905.32 moles} \end{gathered}

Therefore, the number of moles of methane is 905.32 moles

6 0
1 year ago
If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H2SO4(aq)? For your ans
Alik [6]

Question is incomplete, complete question is;

A 34.8 mL solution of H_2SO_4 (aq) of an unknown concentration was titrated with 0.15 M of NaOH(aq).

H_2SO_4(aq)+2NaOH(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H_2SO_4(aq)? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.

Answer:

0.044 M is the molarity of H_2SO_4(aq).

Explanation:

The reaction taking place here is in between acid and base which means that it is a neutralization reaction .

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_1,M_2\text{ and }V_2  are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=?\\V_1=34.8 mL\\n_2=1\\M_2=0.15 M\\V_2=20.4 mL

Putting values in above equation, we get:

2\times M_1\times 34.8 mL=1\times 0.15 M\times 20.4 mL\\\\M_1=\frac{1\times 0.15 M\times 20.4 mL\times 10}{2\times 34.8 mL}=0.044 M

0.044 M is the molarity of H_2SO_4(aq).

4 0
3 years ago
What does our state have that few share?
MissTica

Answer:

which state?

Explanation:

7 0
3 years ago
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